Brief reviewU = Q + W

U, Q, W(We+Wf), Cv( ) Cp, H=U+PV

Initial state A Final state B

State function

Path function

Reversible process

Irreversible process

( )dV

V VQ UC TT

Example 1:

H2O（ l， 0℃）→H2O（ l， 50℃）System: Water

W, Q, U △ ? U = Q + W

For the first case, Q=0, ()V, QV= U=0, is it right? Why?△

Example 2:

Some air in a bicycle pump is compressed so that its volume decreases and its internal energy increases. 

If 25 J of work are done by the person compressing the air, and if 20 J of thermal energy leave the gas through the walls of the pump, what is the increase in the internal energy of the air?

U = Q + W

If psur=constant the whole pump as a system, Qp=△H=0?What happens if we release the pump?  How about the process happen reversibly?

Discussion 1: About the variation of internal energy dU, △U

At constant volume

For ideal gas:

Good approximation for real gases under most conditions

For liquids and solids

So , for all substances without phase transformation

dH, H△

For all substances

Discussion 2: About Cv, Cp

2,mpC a bT cT

2,m / '/pC a b T c T or

Dependence of heat capacity on temperature for real substances

State properties Extensive function

Cv,m, Cp,m Intensive function

Example

Differential scanning calorimetry DSCDifferential thermal analysis DTA

Qualitative and quantitative analysis depending on heat capacity

2.6 Relating Cp and Cv( ) ( )p pV VH UC CT T

( )( ) ( ) p VU PV UT T

( ) ( ) ( )p p VU V UpT T T

( ) ( ) ( ) ( )p pV TU U U VT T V T

( ) ( ) ( )p p pV TU V VC C pV T T

[ ( ) ]( ) pTU Vp V T

For ideal gases( ) 0, TUV

( ) /pV nR pT

p VC C nR

By statistical mechanic CV,m CP,m

Monoatomic 3/2R 5/2R diatomic (or linear molecule ) 5/2R 7/2R polyatomic molecule(or nonlinear molecule) 6/2R=3R 4R

real gas

For any substance other than an ideal gas

For liquids and solids ≈ 0

( ) ( ) ( ) ( )p pV TU U U VT T V T

( , ), ( , )U U T V V V T p

d ( ) d ( ) dV TU UU T VT V

d ( ) d ( ) dp TV VV T pT p

d ( ) d ( ) [( ) d ( ) d ]pV T TU U V VU T T pT V T p

d ( ) ( ) d [( ) ( ) ( ) ]dT T V T pU V U U VU p TV p T V T

=( ) d [( ) ( ) ( ) ]dT V T pU U U Vp Tp T V T

( ) ( ) ( ) ( )p V T pU U U VT T V T

2.7 Gay-Lussac-Joule experiment Q=0 W=0 ΔU=0 Constant energy process

dU=0， dT=0， dV≠0

dVVUdT

TUdU TV )()(

0)(

TVU

Ideal gases

Properties of ideal gases0)(

TVU

ΔH=ΔU + ΔPV =ΔU + nR（ T2-T1）dP

PUdT

TUdU TP )()(

0)(

TPU

U is the function of T only, U(T)

VV TUC )(

( )P PHCT

U, H, Cv, Cp of ideal gases are only the function of T

2.8 Adiabatic processes of ideal gasesdU Q W = 0W Q （ ）

If W>0, U>0, △ △ T>0, T↑

If W<0, U△ <0, △ T<0, T↓

Free expansion: W=0 △U=0, H=0△

If wf=0 dU+pdV=0 dU=CVdT， p=nRT/V

C vdT = - dVVnRT

, ,2 1 1

1 , 2 2

( 1)P m V m

V m

C CT V Vln ln lnT C V V

V

P

mV

mP

CC

CC

,

,

1

2

TT

= 1

2

1 )(

VV

C v,m ln(1

2

TT )=Rln(

2

1

VV

)

T1 V1 γ-1= T2 V2γ-1

1pV K 12TV K 1

3p T K

Adiabatic Process Equation

Comparing with other processesnpV K n= 0 (pressure constant),

1 (isothermal),

γ (adiabatic) ….

A-B Isotherm

1

2

2

12

1

lnlnPPnRT

VVnRTpdVW

2

1

dV

VW p V

2

1

= dV

V

K VV

( )pV K 1 1

2 1= 1 1( )

(1 )K

V V

1 1 2 2pV p V K

2 2 1 1= 1p V pVW

2 1( )1

nR T T

A-C Adiabatic

P

V

C(P2,V2”) B(P2,V2)

A(P1,V1)

Example(a) The pump is operated quickly so the

compression of the air in the cylinder before the valve opens can be considered adiabatic.

At the start of a pump stroke, the pump cylinder contains 4.25 × 10-4 m3 of air at a

pressure of 1.01 × 105 Pa and a temperature of 23 °C. The pressure of air in the dinghy is 1.70 × 105 Pa.  When the valve is about to open, the volume of air in the pump is ?.

 γ for air = 1.4

 (b) Calculate the temperature of the air in the

pump when the valve is about to open.

 V2 = 2.94 x 10-4 m3 

T2 = 344 K

Homework A: P88 3.2, 3.3 3.6 3.7 Y: P21 16 P25 23

Preparation for next class:

The working principle of an refrigerator

A : P 55-69 2.7-2.9

Y:33-47 1.10-1.12